Introduction to Applied Geophysics Exploring the Shallow Subsurface 1st Edition Burger Solutions Man

Answers to Problems: Chapter 2

Analysis of Equation 2.11 suggests that E most directly affects velocity values. The following table supports this assertion given the range of normal variation of the pertinent variables.

1

2.1 V  f  V = 1400m s  10Hz     140m V = 1400m s and f =100Hz;   14m 2.2   450m   45m 2.3 VP  E 1     1 2   1           1 2  0 39  1011 N m 2 1 0 11   2 60g cm 3 1 2 0 11     1  0 11         1 2 VP  0.3471  1011 N m 2 2 2511g cm 3       1 2 VP  1.542  1010 N cm 3 m 2  g 105 dynes N g cm s 2 dyne m 2 104 cm 2       1 2 VP  1 5419  1011 cm 2 s 2       1 2  3 927  105 cm s  3 927  103 m s  3927m s 2.4 VS = 1227 m/s 2.5 VS = 1547 m/s VR = 0.9VS = 1392 m/s

2.6

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Introduction to Applied Geophysics Exploring the Shallow Subsurface 1st Edition Burger Solutions Manual

2 2.7 time  distance velocity timeair  50m 331 5m s  0.151s 151ms timedirect P  50m 2488m s  0.020s  20ms timeRayleigh  50m 0 9  1702m s  0.033s  33ms 2.8 VP = 2829 m/s; VS = 1539 m/s; VRayleigh = 1385 m/s; Vair = 331.5 m/s.

2.9

The drawing below, moving from bottom to top, illustrates successive wavelets emanating from the front of the last wave front. This illustrates in a general way the effect portrayed in Figure 2.18.

3

2.10

This is straightforward because we need only substitute the relevant variables in Equations 2.19–2.21.

4 . . . . . . . . . . . . . . . . . . . . . . . . . . .

sinrfrS  d2 AY and sin i P  d1 AY d1  t1V1P and d2  t1V2S AY = t1V1P sin iP  t1V2 S sinrfrS and sin iP sin rfrS  V1P V2S .

This also is straightforward because we need only substitute the relevant quantities in Figure 2.14 and carry these substitutions through Equations 2.22–2.25. These substitutions are as follows:

(a) The angle of refraction increases as the angle of incidence increases. The angle of refraction always is greater than the angle of incidence until the critical angle for refraction is reached, after which total reflection occurs.

(b) Although the angle of refraction increases as the angle of incidence increases, the angle of refraction always is less than the angle of incidence.

5 2.11

1  iS , 2  rfrP , V1  V1S , and V2  V2P Then sin iS sinrfrP  V1S V2P . 2.12 sin  i P sinrfrP  V1P V2 P sin20Þ sinrfrP  331.5m s 500m s rfrP  31Þ rfrS  12Þ 2.13 rfrP  78Þ rfrS  29Þ 2.14 rfrP  28Þ rfrS 10Þ 2.15 rfrP  16Þ 2.16

2.17

6 icP  sin 1 V1P V2 P        sin 1 1200m s 3800m s       18 4Þ icc  sin 1 V1P V2S        sin 1 1200m s 1900m s       39.2Þ 2.18 tan ic  xcrit 2 h1 and, since sinic  V1 V2 , tan sin 1 V1 V2              xcrit 2 h1 and xcrit = 2h1 tan sin 1 V1 V2            . Introduction to Applied Geophysics Exploring the Shallow Subsurface 1st Edition Burger Solutions Manual Full Download: http://testbanktip.com/download/introduction-to-applied-geophysics-exploring-the-shallow-subsurface-1st-edition-burger-solutions-manual/ Download all pages and all chapters at: TestBankTip.com

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Answers to Problems: Chapter 2

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